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Question

Let $f : D \rightarrow \mathbb R$ and $g : E \rightarrow \mathbb R$ be two uniformly continuous functions with $f(D) \subseteq E$. Show that the composite function $g \circ f : D \rightarrow\mathbb R, x \mapsto g(f(x))$ is uniformly continuous.


Proof

We know that since $f,g$ are continuous

$\Rightarrow\forall \space \epsilon_1\gt0 \space \space \exists \delta_1\gt0 \space$ $\forall x_1,y_1\in D:\space |x_1-y_1|\lt \delta_1\space\space \Rightarrow$ $|f(x_1)-f(y_1)|\lt \epsilon_1$

and

$\forall \space \epsilon_2\gt0 \space \space \exists \delta_2\gt0 \space$ $\forall x_2,y_2\in E:\space |x_2-y_2|\lt \delta_2\space\space\Rightarrow $ $|g(x_2)-g(y_2)|\lt \epsilon_2$.

Since $f(D)\subseteq E$ $\Rightarrow f(x_1),f(x_2)\in E$.

So choosing $x_2=f(x_1)$, $y_2=f(y_2)$ and $\epsilon_1=\delta_2$

We get: $\forall \space \epsilon_1\gt0 \space \space \exists \delta_1\gt0 \space$ $\forall x_1,y_1\in D:\space |x_1-y_1|\lt \delta_1\space\space \Rightarrow$ $|f(x_1)-f(y_1)|\lt \epsilon_1=\delta_2$ $\Rightarrow|g(f(x_1))-g(f(y_2))|\lt \epsilon_2$.

Hence $g \circ f : D \rightarrow\mathbb R, x \mapsto g(f(x))$ is uniformly continuous.


Comments

It would be great if anyone could verify my solution. Any additional proofs or alternative angles would be greatly appreciated too.


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What you had to show was for each $\epsilon>0,$ there exists $\delta>0$ such that $$|x_1-y_1|<\delta \implies |g(f(x_1)-g(f(y_1))|<\epsilon.$$

(Do you see how this is different from your last equation?)

Outline of proof: Let $\epsilon>0,$ since $g$ is uniformly continuous, there exists $\delta'>0$ such that $$|x_2-y_2|<\delta'\implies |g(x_2)-g(y_2)|<\epsilon.$$ Since $f$ is uniformly continuous, there exists $\delta>0$ such that $$|x_1-y_1|<\delta \implies|f(x_1)-f(x_2)|<\delta'\implies |g(f(x_1)-g(f(y_1))|<\epsilon.$$

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  • $\begingroup$ Okay, I think I understand what you're saying. Just updated the last few lines of the proof- is that any better? $\endgroup$ – George Cooper Apr 28 at 14:13
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    $\begingroup$ @GeorgeCooper Yes, but you choose $\epsilon_1 = \delta_2$ not the other way round so be mindful while writing that. $\endgroup$ – Sahiba Arora Apr 28 at 14:20
  • $\begingroup$ Okay, you very much for your help :) $\endgroup$ – George Cooper Apr 28 at 14:28
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Final Answer:

We know that since $f,g$ are continuous

$\Rightarrow\forall \space \epsilon_1\gt0 \space \space \exists \delta_1\gt0 \space$ $\forall x_1,y_1\in D:\space |x_1-y_1|\lt \delta_1\space\space \Rightarrow$ $|f(x_1)-f(y_1)|\lt \epsilon_1$

and

$\forall \space \epsilon_2\gt0 \space \space \exists \delta_2\gt0 \space$ $\forall x_2,y_2\in E:\space |x_2-y_2|\lt \delta_2\space\space\Rightarrow $ $|g(x_2)-g(y_2)|\lt \epsilon_2$.

Since $f(D)\subseteq E$ $\Rightarrow f(x_1),f(x_2)\in E$.

So choosing $x_2=f(x_1)$, $y_2=f(y_2)$ and $\epsilon_1=\delta_2$

We get: $\forall \space \epsilon_1\gt0 \space \space \exists \delta_1\gt0 \space$ $\forall x_1,y_1\in D:\space |x_1-y_1|\lt \delta_1\space\space \Rightarrow$ $|f(x_1)-f(y_1)|\lt \epsilon_1=\delta_2$ $\Rightarrow|g(f(x_1))-g(f(y_2))|\lt \epsilon_2$.

Hence $g \circ f : D \rightarrow\mathbb R, x \mapsto g(f(x))$ is uniformly continuous.

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